Standard Deviation of an Observable

When quantum mechanics predicts the outcome of a measurement it rarely predicts a single definite value. Instead it gives a probability distribution over the eigenvalues of the observable. The standard deviation $\Delta A$ quantifies the spread of that distribution — it is the quantity that appears directly in the Heisenberg uncertainty relation $\Delta x \, \Delta p \geq \hbar/2$ .

Expectation value

Let $A$ be an observable (a Hermitian operator) and $|\psi\rangle$ a normalised state. If $A$ has eigenvalues $a_1, a_2, \ldots$ with corresponding probabilities $p_1, p_2, \ldots$ in the state $|\psi\rangle$ , the expectation value is

$\langle A \rangle = \sum_i p_i\, a_i.$

This is precisely the weighted average of the possible measurement results.

Standard deviation (uncertainty)

The uncertainty $\Delta A$ is the root-mean-square deviation from the mean:

$(\Delta A)^2 = \langle (A - \langle A \rangle)^2 \rangle = \langle A^2 \rangle - \langle A \rangle^2.$

Both identities are equivalent; the second is usually faster to calculate. In practice you compute $\langle A^2 \rangle = \sum_i p_i\, a_i^2$ and then subtract $\langle A \rangle^2$ .

Worked example

Suppose $A$ has two eigenvalues, $a_1 = +1$ (probability $p_1 = 3/4$ ) and $a_2 = -1$ (probability $p_2 = 1/4$ ).

Step 1 — expectation value:

$\langle A \rangle = \tfrac{3}{4}(+1) + \tfrac{1}{4}(-1) = \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}.$

Step 2 — expectation value of $A^2$ :

Since $a_1^2 = a_2^2 = 1$ ,

$\langle A^2 \rangle = \tfrac{3}{4}(1) + \tfrac{1}{4}(1) = 1.$

Step 3 — standard deviation:

$(\Delta A)^2 = \langle A^2 \rangle - \langle A \rangle^2 = 1 - \tfrac{1}{4} = \tfrac{3}{4},$

$\Delta A = \sqrt{\tfrac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866.$

Why it matters

The Heisenberg uncertainty principle bounds the product of two such standard deviations for non-commuting observables. Mastering how to compute $\Delta A$ for a single observable is the essential prerequisite for understanding that principle quantitatively.

Try it

This is a numerical exercise — your code should return a number, not a circuit.

An observable $A$ has eigenvalues $+1$ (probability $3/4$ ) and $-1$ (probability $1/4$ ). Compute $\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}$ .

Run your code to see the quantum state.